# [OPE-L:4347] Re: Problems in Vol. III

john erns (ernst@pipeline.com)
Mon, 10 Mar 1997 23:23:09 -0800 (PST)

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Allin,

I think your OPE-4343 and my OPE-4344 crossed in cyberspace.
At any rate, I have a problem or two with the manner in which
you deal with the IRR for Capital 2.

You wrote:
(snip)

For Cap2, the IRR solves

2000 = sum(t = 1 to 9)[200/(1 + r)^t] + 2200/[(1 + r)^10]

At the end of year 10 the capitalist has \$2200 in hand after
meeting his wage-bill -- his year-10 profit of \$200, as stipulated,
plus the funds to purchase another year's materials, plus the
funds to replace his machine, which have been set aside as
depreciation allowance over the ten years. r = .10

John wonders:

I think you assume that all depreciation funds are recouped only at
the end of the 10 years and, for some reason, are tied up in the
process until the machine wears out. Here's another way of looking
at the matter, given a depreciation charge of 100 per year.

Tied up in Machine Invested in Profit Output
at the beginning of Raw & Aux.
Year of the year Materials

1 1000 1000 200 1300
2 900 1000 200 1300
3 800 1000 200 1300
4 700 1000 200 1300
5 600 1000 200 1300
6 500 1000 200 1300
7 400 1000 200 1300
8 300 1000 200 1300
9 200 1000 200 1300
10 100 1000 200 1300

Totals 5500 10000 2000

Over the 10 years, using the totals we see that the average
rate of profit is 2000/(10000+5500) or about 12.9%. Here
I am treating the capitalist as though he has 10 investments
each of which produce a profit of 200. However, the amounts
invested by "the ten" differ. They are 2000, 1900, 1800 ...
1100. Using this "method", we see that by accelerating the
depreciation process (for example, by overworking machines and
workers in the machine's early years) the capitalist can increase
his rate of return by withdrawing more of his investment in its
early years. This may also be key to understanding "moral
depreciation" but that's another story.

John